Mathematics for the interested outsider

Some Continuous Duals

I really wish I could just say in post titles.

Anyway, I want to investigate the continuous dual of for . That is, we’re excluding the case where either (but not its Hölder conjugate) is infinite. And I say that when is -finite, the space of bounded linear functionals on is isomorphic to .

First, I’m going to define a linear map . Given a function , let be the linear functional defined for any by

It’s clear from the linearity of multiplication and of the integral itself, that this is a linear functional on . Hölder’s inequality itself shows us that not only does the integral on the right exist, but

That is, is a bounded linear functional, and the operator norm is at most the norm of . The extremal case of Hölder’s inequality shows that there is some for which this is an equality, and thus we conclude that . That is, is an isometry of normed vector spaces. Such a mapping has to be an injection, because if then , which implies that .

Now I say that is also a surjection. That is, any bounded linear functional is of the form for some . Indeed, if then we can just pick as a preimage. Thus we may assume that is a nonzero bounded linear functional on , and . We first deal with the case of a totally finite measure space.

In this case, we define a set function on measurable sets by . It’s straightforward to see that is additive. To prove countable additivity, suppose that is the countable disjoint union of a sequence . If we write for the union of through , we find that

Since is continuous, we conclude that , and thus that is a (signed) measure. It should also be clear that implies , and so . The Radon-Nikodym theorem now tells us that there exists an integrable function so that

Linearity tells us that

for simple functions , and also for every , since each such function is the uniform limit of simple functions. We want to show that .

If , then we must show that is essentially bounded. In this case, we find

for every measurable , from which we conclude that a.e., or else we could find some set on which this inequality was violated. Thus .

For other , we can find a measurable with so that . Setting and defining , we find that on , , and so

In the -finite case, we can write as the countable disjoint union of sets with . We let be the union of the first of these sets. We note that for every measurable set , so is a linear functional on of norm at most . The finite case above shows us that there are functions on so that

.

We can define if , and let be the sum of all these . We see that

for every , and since we find that . Then Fatou’s lemma shows us that . Thus the -finite case is true as well.

One case in particular is especially worthy of note: since is Hölder-coonjugate to itself, we find that is isomorphic to its own continuous dual space in the same way that a finite-dimensional inner-product space is isomorphic to its own dual space.

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